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A plane flies at 350 mph in the direction 40° north of east, with a wind blowing at 40 mph in the direction 30° south of east. What is the plane’s drift angle?

5.90°
6.38°
45.89°
46.38°

Sagot :

If the plane flying in direction of 40° north of east at a speed of  350 mph and wind is blowing at 40 mph in the direction 30° south of east. then the drift angle will be 5.9 °.

Let's assume east as x axis and north as y axis.

Breaking velocity of plane into components

350 cos40°  î + 350 sin40°  ĵ

(350 × 0.76604)  î + (350 × 0.64278)  ĵ

268  î + 225  ĵ

Breaking velocity of wind into components

40 cos30°  î - 40 sin30°  ĵ

(40 × 0.86602)  î - (40 × 0.5)  ĵ

34.64  î - 20  ĵ

Now adding velocity of wind to velocity of plane as

The resultant velocity of the plane =  initial velocity of plane + velocity of wind

( 268 + 34.64 )  î + ( 225 - 20 ) ĵ

302.64  î + 205  ĵ

So the final velocity of plane will be 302.64  î + 205  ĵ

Lets assume the angle of final velocity from east be θ

So Tan θ = perpendicular / base

So Tan θ = 205 / 302.64

Tan θ = 0.677

θ = Tan⁻¹(0.677)

θ = 34.12 °

As we know,

Drift angle is the difference of initial and final angle of the object measured from same reference.

Drift angle=| Initial angle of plane from east - final angle of plane from east |

Drift angle= | 40 - 34.12 |

Drift angle= 5.87 ° ≈ 5.9 °

So we used the concepts of vectors to solve the above question, first we find out the the final angle of plane from east and then takes the difference of initial and final angle from east to find the drift angle and that we got as 5.9 °.

Learn more about Vectors here:

brainly.com/question/25705666

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